See Problem 33.

By the above discussion, an affine scheme \(\operatorname{Spec}(R)\) is reduced and irreducible if and only if \(R\) is a domain. If \(X\) is a reduced (or irreducible, respectively) scheme, then so is every non-empty open subscheme. Together we obtain (i) \(\Rightarrow \) (ii). The implication (ii) \(\Rightarrow \) (iii) is trivial. Regarding (iii) \(\Rightarrow \) (i), we have already seen that (iii) implies that \(X\) is reduced.

It remains to show that (iii) implies that \(X\) is irreducible. Otherwise, there exist non-empty open subsets \(U, U' \subseteq X\) with empty intersection. By shrinking \(U\) and \(U'\), if necessary, we may assume that \(U\) and \(U'\) are affine open subschemes. Then \(U \cup U'\) is a non-irreducible affine open subscheme, so \(\Gamma (U\cup U', {\mathscr O}_X)\) is not a domain.

Because of a lack of time we skipped this proof in class. It is easy to check that \(I_{V({\mathfrak a})} = {\mathfrak a}\). Therefore it is enough to show that for every closed subscheme \(Z\), we have \(Z = V(I_Z)\).

The ring homomorphism \(A\to \Gamma (Z, {\mathscr O}_Z)\) factors through \(A/I_Z\), and we want to show that the natural morphism \(Z\to \operatorname{Spec}(A/I_Z)\) is an isomorphism. Replacing \(A\) by \(A/{\mathscr I}\), we may assume that \(\varphi \colon A\to \Gamma (Z, {\mathscr O}_Z)\) is injective (and then want to show that \(Z\cong \operatorname{Spec}(A)\)).

(I) Let us show that the map \(Z\to \operatorname{Spec}(A)\) is a homeomorphism. We know already that it is injective and closed; it remains to show its surjectivity. We will show that \(Z\) is not contained in a proper closed subset of \(\operatorname{Spec}(A)\). Let \(s\in A\) with \(Z \subseteq V(s)\); we will show that then necessarily \(s\) is nilpotent, i.e., \(V(s) = \operatorname{Spec}(A)\).

In view of the injectivity of \(\varphi \), it is enough to check that \(\varphi (s)\) is nilpotent. Let \(Z = \bigcup _i V_i\) be a finite affine open cover (\(Z\) is quasi-compact since it is closed in the quasi-compact affine scheme \(\operatorname{Spec}(A)\)). Then \(\Gamma (Z, {\mathscr O}_Z)\) injects into the product \(\prod _i \Gamma (V_i, {\mathscr O}_Z)\), so it is enough to show that all restrictions \(s_{|V_i}\) are nilpotent. But \(V_i \subseteq Z \subseteq V(s)\) implies \(V_i = V_{V_i}(s_{|V_i})\), so \(s_{|V_i}\in \Gamma (V_i, {\mathscr O}_Z)\) is indeed nilpotent.

(II) To conclude the proof, we show that \(Z=\operatorname{Spec}(R)=X\) as schemes, i.e., the sheaf homomorphism \({\mathscr O}_X\to {\mathscr O}_Z\) is an isomorphism (we identify \(Z=X\) as topological spaces in view of Step (I)). This sheaf homomorphism is surjective by assumption, and it remains to show the injectivity. We check this on the stalks.

So let \({\mathfrak p}\subset R\) be a prime ideal (i.e., a point of \(X\)), and consider the ring homomorphism \(R_{{\mathfrak p}}={\mathscr O}_{X, {\mathfrak p}}\to {\mathscr O}_{Z, {\mathfrak p}}\). It is then enough to show that for all \(g\in A\) with \(\frac{g}{1}\mapsto 0\in {\mathscr O}_{Z, {\mathfrak p}}\) we have \(g=0\).

So fix \(g\) with this property. Let \(V \subseteq Z\) be an affine open neighborhood of \({\mathfrak p}\) such that \(\varphi (g)_{|V} = 0\). Choose an affine open cover \(Z = \bigcup _{i=1}^n V_i\) with \(V_1 = V\).

Let \(s\in A\) with \(D(s) \subseteq V\).

Claim. There exists \(N\ge 0\) such that \(\varphi (s^Ng)=0\in \Gamma (Z, {\mathscr O}_Z)\).

We can check this on each \(V_i\) separately. By assumption \(\varphi (g_{|V}) = 0\); it remains to handle the case \(i {\gt} 1\). Since \(D_{V_i}(\varphi (s)_{|V_i}) = D(s)\cap V_i \subseteq V\cap V_i\), we have \(g_{|D_{V_i}(\varphi (s)_{|V_i})} = 0\), i.e., the image of \(\varphi (g)\) in the localization \(\Gamma (V_i, {\mathscr O}_Z)_{\varphi (s)_{|V_i}}\) is \(=0\). Thus \((\varphi (s)_{|V_i})^{N_i} \varphi (g)_{|V_i} = 0\) for some \(N_i\).

The claim is proved. Since \(\varphi \) is injective, it follows that \(s^N g = 0\), so the image of \(g\) in \(A_s\) is \(=0\). A fortiori, \(g\) maps to \(0\) in \(A_{{\mathfrak p}}\). This is what remained to show.

If \(X=\operatorname{Spec}(R)\) is affine and \(Z = V({\mathfrak a})\) as a set, for an ideal \({\mathfrak a}\subseteq R\), then the scheme \(V(\sqrt{{\mathfrak a}})\) is the unique reduced closed subscheme of \(X\) with underlying set \(Z\) (recall, cf. Proposition 2.9, that \(V({\mathfrak a})\) and \(V({\mathfrak b})\) have the same underlying closed subset, if and only if the radicals of \({\mathfrak a}\) and \({\mathfrak b}\) are equal; the quotient of a ring by an ideal is reduced if and only if the ideal is a radical ideal).

In the general case, in view of the uniqueness statement, we may obtain the desired closed subscheme by using gluing of schemes.

Everything is clear by the above except for the claim about morphisms \(T\to X\) from reduced schemes to \(X\) which can be reduced to the case where \(X=\operatorname{Spec}(R)\) is affine. In that case, \(X_{\rm red} = \operatorname{Spec}(R/{\rm nil}(R))\) is the affine scheme attached to the quotient of \(R\) by its nil radical, and the claim is checked easily.

Let \(\varphi \colon A\to B\) be a ring homomorphism. Then the natural homomorphism \(B\otimes _A B\to B\), \(b\otimes b'\mapsto bb'\), is surjective. This implies the proposition.

The question is local on the target, so we may assume that \(S\) is affine. Let \(X = \bigcup _i U_i\) be an affine open cover. Then \(\Delta \) factors as \(X\to \bigcup _i (U_i\times _S U_i) \to X\times _SX\), and the second morphism is an open immersion. Thus it is enough to check that the first morphism is a closed immersion. This can be checked locally on the target, i.e., it remains to observe that (by Proposition 7.16) for every \(i\), \(U_i\to U_i\times _S U_i\) is a closed immersion.

Let \(E = {\rm Eq}(f,g)\) be the equalizer of \(f\) and \(g\), i.e., the scheme defined by the following fiber product diagram

The morphism in the lower row is the morphism \(X\to Y\times _SY\) obtained by using the universal property of the fiber product \(Y\times _SY\) for the two morphisms \(f\) and \(g\). (Since \(\Delta \) is always an immersion, the same is true for the morphism \(E\to X\); here we use that the property of being an immersion is stable under base change.)

The universal property of this particular fiber product translates into the following universal property for the scheme \(E\). A morphism \(h\colon T\to X\) factors through the projection \(E\to X\) if and only if the compositions \(f\circ h\) and \(g\circ h\) are equal.

This shows that the inclusion \(U\to X\) factors through \(E\). We want to show that the identity morphism \(X\to X\) also factors through \(E\) (i.e., that \(E=X\)). Since \(Y/S\) is separated by assumption, in the situation at hand \(\Delta \) and hence the morphism \(E\to X\) are closed immersions. Since \(U \subseteq E\) and \(U\) is dense in \(X\), the underlying topological space of \(E\) is equal to \(X\). Since \(X\) is reduced, this implies that \(E=X\), as desired.

We have a cartesian diagram

Since the lower horizontal arrow is a closed immersion by assumption, so is the upper one. Now \(U\times _SV\) is affine, and every closed subscheme of an affine scheme is itself affine by Theorem 7.9.

The implications (i) \(\Rightarrow \) (ii) \(\Rightarrow \) (iii) follow from what we have already shown. It remains to show (iii) \(\Rightarrow \) (i), so assume that \(X = \bigcup _i U_i\) is an affine open cover as in (iii). To show that \(\Delta \colon X\to X\times _SX\) is a closed immersion, we may work locally on the target. We have \(X\times _SX = \bigcup _{i,j} U_i\times _S U_j\). Since \(\Delta ^{-1}(U_i\times _S U_j) = U_i\cap U_j\), and the morphisms

are closed immersions by the assumptions in (iii), the result follows.

Since being separated is stable under base change, it is enough to consider the case \(S=\operatorname{Spec}(\mathbb {Z})\). In this case, the result follows from the previous proposition, applied to the standard cover \(\mathbb {P}^n_{\mathbb {Z}} = \bigcup _i D_+(X_i)\).